An Electrical Question

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Gauge

Aquarium Advice FINatic
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I'm going to be building a new canopy for my tank and I want to put some moon lights on it. I've looked and the prices for buying them commercially seem really ridiculous considering that I can buy a "super bright" LED for like a dollar at a store nearby. I know that the LEDs I would use are 5 volt LEDs. Does that mean that all I need is one of those multi-use AC adapters you can buy at like Wal-Mart? Basically, just set it to 5 volts, cut the connector off the end and solder it to the LED(s)?

Or do I need to do something more involved like using a resistor to control the current? I'm a computer science major, not an electrical engineering major. I'm an electrical retard.
 
You got the right idea, i used three LED's on my 30 gal tank...the only thing you will need are resistors so you don't blow out the LED's. All I did was wire all the leds and resistors together and connected them to a 12 volt transformer (like the ones you mentioned) Those DIY sites should help alot. :wink:
 
Thanks, guys. I actually busted out my father's old college electronics books to figure it out mathematically. I had the idea that any "appliance" (i.e. any electricity-using device) automatically drew the correct ammount of current, because I was once told this. I took this too literally because the "appliances" they were referring to had the necessary transformers and resistors in them.

Looking at this ancient book of my father's, I think Ohm's Law was the most applicable piece of information I found...

E=IR

Thus, if it's a 5 volt LED, I should be getting a 5 volt transformer (AC adapter) and then figuring the type of resistor I need by this variance of Ohm's Law...

R=E/R (where E is voltage and I is current)

Does this sound accurate? And how would you know what wattage the LEDs require? Is that guaranteed to be given information or not?



Thanks for all the help. I'll be looking at those DIY things, but I want to make sure my logic is correct regardless. This has become like a little puzzle or game for me. :)
 
Almost,

You are on the right track, but the object with LEDs is that they are not resistive for all intents and purposes so the current in their circuit must be limited to their foward current. Typically this is 20mA (MilliAmps)

So here is the formula you should use to determine the resistor you should use ( and there is a little play here). But first you will need to know the LED's forward Bias voltage voltage - there are typlically only a few of these and they vary by the color of the LED:

Red: 1.7 Volts
Green 2.5 Volts
Blue: 3.9 volts

The Forumla is a Variation of Ohms Law (BTW I can tell your Dad's book is old as we typically do not use E - Electromotive Force, but instead use V - Volts).

Resistor's Value = Supply Volts - LED Forward Volts / LED Current Max

By using this formula you limit the current in the LED pass circuit and also appropriately adjust the voltage. So for a 5 volt supply and a BLUE LED:

R = 5 - 3.9 / .02 Amps = 55 Ohms

You could use a 12V supply as well:

R = 12 - 3.9 / .02 Amps = 405 Ohms

The thing to watch for on the supply is how many mA it is rated at. recall that each resistor/diode branch uses 20mA.


HTH

I've posted in DIY forum under '5 LED Moonlight' a diagram and a link to get a really cheap power pack - I think $3

Tom
 
Wow, thanks for all the help, guys...

Btw, that R=E/R was a typo... I meant R=E/I, of course. I think that the only question that I've got so far is this - What are "forward volts"?

I'm sure that I can follow your awesome directions (thanks again, btw), but I'm curious to understand it a little better. :)


For those of you who want to see the thread TheMadNucleus was talking about here is a link
 
Hi Guage,

np on the direction. Good to see some one so interested.

As for foward volts, it's not so complicated. First let me expand a little on the term, it is actually a transistor rating and is in short form. The full term is the valtage required to forward bias a transistor.

Transistors are, for lack of a better term, electronic (electron) switches. They work with multiple layers of silicon and other semiconductors. LED's are a type of transistor. Rather then switching electrons they pass electrons, but when they do they give off photons as a byproduct of this electron conductivity.

So that being said - what is forward voltage? Forward voltage is the voltage required to get the transistor (or LED) to move electrons from one junction to another (in the case of an LED from the negative connection to the positive connection). For example for a Red LED this is about 1.7 volts. So if you apply 1.4 volts - no electrons will pass through the LED (semiconductors) and it won't light, but if you apply 1.8 volts then a flood of electrons will begin to cross through the semiconductor and Viola - red light.

This is why LED's are not particularly dimmable. Because they are sort of sensitive to this bias - varying the voltage doesn't really affect the forward bias to much in increments (a little but not worht using as a dimmer). I.e. the thing is sort of on or off.

So how do you dim them? Well - you use a simple electronic pulse generator which turns the LED on and off so quickly that your eye can not detect that it is flashing. Since it is a semiconductor - it can do this (a regular light bulb could not). Then in this pulse generator you allow it to adjust the width if the pulse from on to just 10% of the time to on for pehaps 90% of the time. Viola - an LED dimmer.

Well - guess I'm rambling - but you asked :)

Lata'

Tom
 
So, the forward voltage is actually lost as the semiconductor gap is crossed? That's why you subtract it out of the equation for the correct resistor?

Also, I plan on running probably five LEDs in parallel. If I want to have just one resistor (since all the LEDs will be the same type), then I would use the following math...

R = V / I
R = (Applied Voltage - Forward Voltage) / (5 x Amperage)
R = (5v - 3.9v) / (5 x 20mA)
R = 1.1v / 0.1A
R = 11 ohms

Is this correct? I seem to remember reading that voltage is constant in a parallel circuit.
 
That looks right to me, but I'm no math guru. Voltage does remain constant in a parallel circuit as long as the wiring and power supply are capable of supplying the needed power. Voltage drops in a series circuit. For instance, you can hook 10 12v bulbs up in series and connect them to 120v...they will light just fine.
 
A friend of mine has taken interest... He, unlike me, wants to do a combination of white and blue LEDs. Is there a way to have them on the same parallel circuit? I've not heard of a way to drop the voltage yet... only the current.
 
haha yea metku is a good place for info too. japala is a good guy. he knows what he is doing.
 
Actually I had seen both those pages already. :p

I think that the misunderstanding I'm having involves LEDs themselves. My understanding is that the resistances of the "appliance" sets the amperage for itself. Thus, the resistance of a lightbulb naturally lets the correct amperage through. And a lower wattage bulb running on the same voltage will have a higher resistance to it.

This understanding, however, is contradicted with LEDs, though. Do they not have a resistance of their own? If there is no resistance, then there is no voltage across the appliance. Yet, they require a voltage?

How are they different?
 
Yes there is still a voltage across the LED. Someone can correct me if i'm wrong, but you can think of the LED as almost a wire that requires a certain voltage. In order to get the specified voltage, you must have an inline resistor to reduce the voltage. A light bulb is a resistor. The tungsten filament just happens to be a resistor that gives off light, where as a regular resistor tends to just disperse the energy as heat.

The LED translates nearly 100% (?) of the voltage to light without heat. Well, there si some heat, but very minimal. Where's TheMadNucleus? He's got this stuff down. I just do mechanical and computer systems :roll:

Jim
 
Sorry Folks,

Work got in the way.

Hey so far everything here looks ok. A few comments on your thoughts on parallel with one resistor and the resistance of LED's. They do have some level of resistance (though this is not fixed like most appliances - that is a character of diodes) but since they are current sensitive the circuits are sort of designed to be little current limitors. But you can find the resistance at a given operating voltage using ohms law as well:

for a blue LED:

Max current = 20ma at 3.9 volts so resistance = 3.9 / .020 Amps = 195 ohms

This sort makes sense as the formula I gave you earlier (Vs - Vf)/Iled is a combination of these, note 5-3.9/.020 = 55 ohms

so add the ohms 195 + 55 = 250 ohms

using vanilla version of Ohm's V=250 * .02 and we're back at our 5 volts.

So there are two objects in an LED circuit: 1.) supply the correct volatage across the LED and 2.) Limit the current in the whole circuit to 20 ma. The LED current Limiting resistor does this job.

On running with one resistor for parallel LED's this is not a good idea. This is why:
If you use a resitor in parallel you can specifically mix and match colored LED as each has it's own requirements and also you are not fixed in the number of LED's based on the resistor you chose. Other than that - you can in fact add up the current requirements for as many LED's you have, and use the same formula to arrive at the single Load/current limiting resistor in your circuit.

For example:

5v - 3.9v/ (.02 + .02 + .02...nLED)

See chart

BTW - the law that shows that the 60ma going into node one gets split to 40ma and 20 ma is Kirchoff's current law: The sum of the currents entring a node is equal to the sum of the currents leaving the node :)


HTH

Tom
 
Thanks Tom. :)

Electricity is confusing... and painful. I think I understand, though. The reason that you subtract the voltage of the LED is so that the voltage that you're removing with the resistor is equal to the difference between the applied voltage and the desired voltage... thus yielding the desired voltage. Correct?

And, just to clarify, the resistance of an LED is proportional to the voltage going through it?
 
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